What does $\sum\limits_{n=1}^{N-1} \frac{1}{n} - \sum_{n=3}^{N+1} \frac{1}{n} $ simplify to?

Question

A solution to one of the exercises in my text states:

$$\sum\limits_{n=1}^{N-1} \frac{1}{n} - \sum_{n=3}^{N+1} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} - \frac{1}{N} - \frac{1}{N+1}$$

I have no idea how to get the right hand side of the above equation.

I do realize that the left hand side involves finite sums of harmonic series which as far as I know, there is no closed form solution. So I am really puzzled as to the manipulations to be done to arrive at the right hand side. Please tell me how to get the left right hand side.

Judging by how my solution states it, it seems like this is a standard result. So please also let me know what are the properties of summation that are being used so that I can solve similar questions myself in the future.

Answer 1

Note that $$\sum_{n=1}^{N-1}\frac 1n=\frac 11+\frac 12+\color{red}{\frac{1}{3}+\frac14+\cdots+\frac{1}{N-1}}$$ and that $$\sum_{n=3}^{N+1}\frac 1n=\color{red}{\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{N-1}}+\frac{1}{N}+\frac{1}{N+1}$$

Answer 2

$\sum_{n=1}^{N-1}\frac{1}{n}-\sum_{n=3}^{N+1}\frac{1}{n}$

$=\frac{1}{1}+\frac{1}{2}+\sum_{n=3}^{N-1}\frac{1}{n}-\sum_{n=3}^{N-1}\frac{1}{n}-\frac{1}{N}-\frac{1}{N+1}$

$=\frac{1}{1}+\frac{1}{2}-\frac{1}{N}-\frac{1}{N+1}$

As for the harmonic series, this isn't a harmonic series per se, rather it is a difference of finite harmonic series with just different boundaries of summation, which should clear up why such a closed form is possible.

Answer 3

First thing to notice:

$$\sum_{n=1}^{N-1}\frac 1n=\frac 11+\frac 12+\cdots+\frac{1}{N-1}$$ and that $$\sum_{n=3}^{N+1}\frac 1n=\frac{1}{3}+\cdots+\frac{1}{N}+\frac{1}{N+1}$$

So what we get:

$$\left(\sum_{n=1}^{N-1}\frac 1n\right)-\left(\sum_{n=3}^{N+1}\frac 1n\right)=\frac{3}{2}-\frac{2N+1}{N^2+N}$$